On an inclined plane of inclination `30^(@)`, a ball is thrown at angle of `60^(@)` with the horizontal from the foot of the incline with a velocity of `10sqrt(3)ms^(-1)`. Then the ball will hit the inclined plane in

Options

A. `1s`
B. `2s`
C. `2sqrt(3)s`
D. `4sqrt(3)s`

Explanation

To solve the problem of how long the ball will hit the inclined plane, we will follow these steps:

### Step 1: Understand the Problem
We have a ball thrown from the foot of an inclined plane at an angle of \(60^\circ\) with the horizontal. The incline itself is at an angle of \(30^\circ\). The initial velocity of the ball is \(10\sqrt{3} \, \text{m/s}\). We need to find the time it takes for the ball to hit the inclined plane.

### Step 2: Resolve the Initial Velocity
The initial velocity \(u\) can be resolved into two components:
- The horizontal component \(u_x = u \cos(60^\circ)\)
- The vertical component \(u_y = u \sin(60^\circ)\)

Calculating these components:
- \(u_x = 10\sqrt{3} \cos(60^\circ) = 10\sqrt{3} \cdot \frac{1}{2} = 5\sqrt{3} \, \text{m/s}\)
- \(u_y = 10\sqrt{3} \sin(60^\circ) = 10\sqrt{3} \cdot \frac{\sqrt{3}}{2} = 15 \, \text{m/s}\)

### Step 3: Determine the Equations of Motion
The equations of motion for the ball can be described as:
- Horizontal motion: \(x = u_x \cdot t\)
- Vertical motion: \(y = u_y \cdot t - \frac{1}{2} g t^2\)

Where \(g = 10 \, \text{m/s}^2\) is the acceleration due to gravity.

### Step 4: Find the Equation of the Inclined Plane
The equation of the inclined plane can be derived from its angle:
- The slope of the incline is given by \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\).
- Thus, the equation of the inclined plane is \(y = \frac{1}{\sqrt{3}} x\).

### Step 5: Substitute the Motion Equations into the Inclined Plane Equation
We substitute \(x\) and \(y\) from the motion equations into the inclined plane equation:
\[ u_y \cdot t - \frac{1}{2} g t^2 = \frac{1}{\sqrt{3}} (u_x \cdot t) \]
Substituting the values:
\[ 15t - \frac{1}{2} \cdot 10 t^2 = \frac{1}{\sqrt{3}} (5\sqrt{3} t) \]
This simplifies to:
\[ 15t - 5t^2 = 5t \]
Rearranging gives:
\[ 10t - 5t^2 = 0 \]
Factoring out \(t\):
\[ t(10 - 5t) = 0 \]
This gives us two solutions:
1. \(t = 0\) (the time of projection)
2. \(10 - 5t = 0 \Rightarrow t = 2 \, \text{s}\)

### Step 6: Conclusion
The ball will hit the inclined plane after \(t = 2 \, \text{s}\).

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